Mathematics · Oct / Nov 2023 · Paper 31 · Question 10

1

π

=

sin 4

2

y

y

=

4

24

d

x

y =

8sin 4 cos4

y

y

d

d

x

d

y

1

=

8sin 4 cos4

y

y

→

=

d

y

d

x

8sin 4 cos4

y

y

d

y

1

=

d

x

8

x

(1

−

x

)

d

y

1

=

d

x

2

1

16

64

−

x

−

2

(d)(i)

1

x =

2

(ii)

d

y

1

x =

d

4

Notes

(a)

1

M1:

Attempts to substitute in

x =

and rearranges to find an exact value for y. Look for

4

1

→

y

=

M

arcsin

±

k

π

or k(30°) allowing errors in dividing by the 4, or equivalents criteria

2

via a double angle identity, correct up to sign error.

π Ignore extra solutions outside the domain.

A1:

24

(b)

d

x

=

A

sin 4 cos4

y

y

A

sin8

y oe Accept

M1:

Attempts to differentiate achieving a form

or

d

y

d

y

. The d

x

alternative forms e.g. via implicit differentiation

1

=

A

sin 4 cos4 d

y

y x

y may be

d

missing, or labelled d

y

x for this mark.

d

oe eg d

Coefficients must be simplified. Must include the d

x

x

A1:

y =

8sin 4 cos4

y

y

y =

4sin8

y

d

d

(c)

Uses d

y

d

x

M1:

= ÷

1

(allow for the reciprocal of their answer to (b) if left hand side is omitted).

d

x

d

y

Variables must be consistent.

to write d

y

x or d

x

M1:

Attempts to use sin 4y

= ±

x

and cos4

y

=

±

1

−

x

y in terms of x only

d

d

with no trig terms.

d

y

1

=

d

x

A1:

2

and isw after correct answer. Allow with terms under the square root

1

16

−

64

x

−

2

reversed.

(d) Note to score marks in (d) they must have a derivative which does have a minimum and is of a

completed square form (the r may be inside the bracket) or in some other way clearly uses their (c)

(e.g. minimising their quadratic).

) 1

(

(i) B1ft:

x =

ft their −s provided their q>0, their r <0 and their x is in the range 0

2

d

y

1

(ii) B1ft:

=

ft their q provided it is positive and their r was negative.

d

x

4

(b)

1

d

y

1

1

1

1

−

x

=

sin 4

2

y

y

=

arcsin

x

=

×

x

2

Alt

4

d

x

4

(

)

2

2

1

−

x

d

x

=

8

x

1

−

x

d

y

(c)

d

x

d

y

1

=

8sin 4 cos4

y

y

→

=

d

y

d

x

8sin 4 cos4

y

y

d

y

1

=

d

x

8

(1

)

x

−

x

d

y

1

=

d

x

1

2

16

−

64

x

−

2

Notes

Alt by finding y first. Mark (b) and (c) together via such approaches. But note that part (c) says

“Hence” and the reciprocal law must have been used at some stage to score full marks.

(b)

d

y

1

M1: Makes y the subject and differentiates to reach form

=

A

×

...

where … is a

d

x

(

)

2

1

−

x

function of x

A1: For d

x

=

8

x

1

−

x

oe with square roots combined. Coefficients must have been gathered.

d

y

(c)

M1: For use of the reciprocal rule of derivatives evidenced in the working – allow for it being used in

(b) to find d

y from d

x

y

x .

d

d

M1: Award for the correct procedure of having made y the subject and differentiating to reach the

d

y

1

1

−

correct form for the answer, ie

=

A

×

Bx

2

d

x

(

)

2

1

−

x

A1: As main scheme.

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